【蓝桥备赛】妮妮的月饼工厂——二分查找

题目链接

妮妮的月饼工厂

个人思路

通过二分查找，寻找满足条件的高度，判定标准是当我们选择mid高度时，我们可以切出的月饼个数是否满足题目要求的 K 个。

	static boolean check(long mid) {
        if (mid == 0) return false;
        long res= 0;
        for (int i = 0; i < n; ++i) {
            res += arr[i] / mid;
        }
        return res >= k;
    }

当然，此处在二分查找的时候，因为我的左边界是从 0 开始的，且是左闭右闭区间，所以在处理的时候需要判断 0 的情况直接返回 false。

参考代码

Java

import java.io.*;

public class Main {
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static int n;
    static long k;
    static long[] arr;
    static boolean check(long mid) {
        if (mid == 0) return false;
        long res= 0;
        for (int i = 0; i < n; ++i) {
            res += arr[i] / mid;
        }
        return res >= k;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner();
        n = sc.nextInt();
        k = sc.nextLong();
        arr = new long[n];
        long l = 0, r = 0;
        for (int i = 0; i < n; ++i) {
            arr[i] = sc.nextLong();
            r= Math.max(arr[i], r);
        }
        while (l <= r) {
            long mid = (r - l) / 2 + l;
            if (check(mid)) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        if (check(r)) {
            out.println(r);
        } else {
            out.println(-1);
        }
        out.flush();
    }
}
// Java 快读
class Scanner {
    static StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    public int nextInt() {
        try {
            st.nextToken();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
        return (int) st.nval;
    }
    public long nextLong() {
        try {
            st.nextToken();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
        return (long) st.nval;
    }
}

C/C++

#include
using namespace std;
typedef long long ll;
const int N = 1e5 + 3;
int n;
ll k, arr[N];

int check(ll mid)
{
    if(mid <= 0) return 0;
    ll cnt = 0;
    for(int i = 1; i <= n; ++i) cnt += arr[i] / mid;
    return cnt >= k;
}

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> k;
    ll l = 0, r = 0;
    for(int i = 1; i <= n; ++i) cin >> arr[i], r = max(r, arr[i]);
    while(l <= r)
    {
        ll mid = (r - l) / 2 + l;
        if(check(mid)) l = mid + 1;
        else r = mid - 1;
    }
    if(check(r)) cout << r;
    else cout << -1;
}